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7.7. Penulangan Balok Portal
7.7.1. Perhitungan Tulangan Lentur Balok Portal Memanjang
Data perencanaan : h = 500 mm
b = 300 mm p = 40 mm
fy = 360 Mpa f’c = 25 MPa
Ø
t
= 19 mm Ø
s
= 10 mm d = h - p - Ø
s
- ½.Ø
t
= 500 – 40 – 10 - ½.19 = 440,5 mm
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
85 ,
360 25
. 85
, = 0,031
ρ
max
= 0,75 . ρb
= 0,75 . 0,031 =
0,0232 ρ
min
= 0038
, 360
4 ,
1 4
, 1
= =
fy
Daerah Tumpuan Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 229 :
Mu = 3747,46 kgm = 3,747 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 747
, 3
7
× = 4,68 × 10
7
Nmm Rn =
804 ,
440,5 300
10 4,68
d .
b Mn
2 7
2
= ×
× =
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m = 17
25 0,85
360 c
0,85.f fy
= ×
= ρ =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
804 ,
17 2
1 1
17 1
= 0,0023 ρ
ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ
min
= 0,0038 As perlu =
ρ. b . d = 0,0038× 300 × 440,5
= 502,17 mm
2
Digunakan tulangan D 19 n
= 385
, 283
17 ,
502 19
. 4
1 perlu
As
2
= π
= 1,77 ≈ 2 tulangan
As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As ada = 2 × 283,385 = 566,77 mm
2
As’ As………………….aman Ok
Jadi dipakai tulangan 2 D 19 Daerah Lapangan
Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 229.
Mu = 2308,61 kgm = 2,308 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 308
, 2
7
× = 2,885 × 10
7
Nmm Rn =
495 ,
440,5 300
10 2,885
d .
b Mn
2 7
2
= ×
× =
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m = 17
25 0,85
360 c
0,85.f fy
= ×
= ρ =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
495 ,
17 2
1 1
17 1
= 0,0014 ρ
ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ
min
= 0,0038 As perlu =
ρ. b . d = 0,0038 × 300 × 440,5
= 502,17 mm
2
Digunakan tulangan D 19 n
= 385
, 283
17 ,
502 19
. 4
1 perlu
As
2
= π
= 1,77 ≈ 2 tulangan
As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As ada = 2 × 283,385 = 566,77 mm
2
As’ As………………….aman Ok
Jadi dipakai tulangan 2 D 19
7.7.2. Perhitungan Tulangan Geser Portal Memanjang Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 229:
Vu = 5546,31 kg = 55463,1 N
f’c = 25 Mpa
fy = 360 Mpa
d = 440,5
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Vc = 1 6 .
c f .b .d
= 1 6 . 25 .300.440,5
= 110125 N φ Vc = 0,75 .110125 = 82593,75 N
½ Ø Vc = 0,5 . 82593,75 = 41296,87 N 3
φ Vc = 3 . 82593,75 = 247781,25 N Syarat tulangan geser : ½ Ø Vc Vu Ø Vc
: 41296,87 N 55463,1 N 82593,75 N Jadi diperlukan tulangan geser
Ø Vs = Vu – ½ Ø Vc
= 55463,1 - 41296,87 = 14166,23 N
Vs perlu =
75 ,
14166,23 75
, =
Vs φ
= 18888,31 N Av
= 2 . ¼ π 10
2
= 2 . ¼ . 3,14 . 100 = 157 mm
2
S =
12 ,
1318 18888,31
5 ,
440 .
360 .
157 perlu
Vs d
. fy
. Av
= =
mm S max = d2 =
2 5
, 440
= 220,25 mm
Jadi dipakai sengkang dengan tulangan Ø 10 – 200 mm
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50
300
2 D19
Ø10 -200 2 D19
50
300
2 D19
Ø10 -200 2 D19
Tul. Tumpuan Tul. Lapangan
Potongan balok portal memanjang
7.7.3. Perhitungan Tulangan Lentur Balok Portal Melintang
Data perencanaan : h = 500 mm
b = 300 mm p = 40 mm
fy = 360 Mpa f’c = 25 MPa
Ø
t
= 19 mm Ø
s
= 10 mm d = h - p - Ø
s
- ½.Ø
t
= 500 – 40 – 10 - ½.19 = 440,5 mm
ρb = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ + fy
600 600
fy c.
β 0,85.f
= ⎟
⎠ ⎞
⎜ ⎝
⎛ + 360
600 600
85 ,
360 25
. 85
, = 0,031
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ρ
max
= 0,75 . ρb
= 0,75 . 0,031 =
0,0232 ρ
min
= 0038
, 360
4 ,
1 4
, 1
= =
fy
Daerah Tumpuan Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278.
Mu = 13309,39 kgm = 13,309 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 309
, 13
7
× = 16,64 × 10
7
Nmm Rn =
85 ,
2 440,5
300 10
16,64 d
. b
Mn
2 7
2
= ×
× =
m = 17
25 0,85
360 c
0,85.f fy
= ×
= ρ =
⎟⎟ ⎠
⎞ ⎜⎜
⎝ ⎛
− −
fy 2.m.Rn
1 1
m 1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
85 ,
2 17
2 1
1 17
1 = 0,0085
ρ ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ = 0,0085 As perlu =
ρ. b . d = 0,0085 × 300 × 440,5
= 1123,27 mm
2
Digunakan tulangan D 19 n
= 385
, 283
27 ,
1123 19
. 4
1 perlu
As
2
= π
= 3,964
≈ 4 tulangan
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As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As = 4 × 283,385 = 1133,54 mm
2
As’ As………………….aman Ok
Jadi dipakai tulangan 4 D 19 Daerah Lapangan
Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278.
Mu = 11375,55 = 11,375 × 10
7
Nmm Mn
= φ
Mu =
8 ,
10 375
, 11
7
× = 14,22 × 10
7
Nmm Rn =
44 ,
2 440,5
300 10
14,22 d
. b
Mn
2 7
2
= ×
× =
m = 17
25 0,85
360 c
0,85.f fy
= ×
=
ρ = ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ −
− fy
2.m.Rn 1
1 m
1
= ⎟⎟
⎠ ⎞
⎜⎜ ⎝
⎛ ×
× −
− 360
44 ,
2 17
2 1
1 17
1 = 0,0072
ρ ρ
min
ρ ρ
max
→ dipakai tulangan tunggal Digunakan
ρ
perlu
= 0,0072 As perlu =
ρ. b . d = 0,0072 × 300 × 440,5
= 951,48 mm
2
Digunakan tulangan D 19 n
= 385
, 283
48 ,
951 19
. 4
1 perlu
As
2
= π
= 3,357 ≈ 4 tulangan
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As’ =
2
19 .
4 1 π
=
2
19 .
14 ,
3 4
1 = 283,385 mm
As ada = 4 × 283,385 = 1133,54 mm
2
As’ As………………….aman Ok
Jadi dipakai tulangan 4 D 19 7.7.4.
Perhitungan Tulangan Geser Balok Portal Melintang Dari Perhitungan SAP 2000 diperoleh momen terbesar pada batang nomor 278:
Vu = 13607,10
kg = 136071,0 N
f’c = 25 Mpa
fy = 360 Mpa
d = 440,5
Vc = 1 6 .
c f .b .d
= 1 6 . 25 .300.440,5 = 110125 N
φ Vc = 0,75 . 110125 = 82593,75 N ½ Ø Vc = 0,5 . 82593,75 = 41296,87 N
3 φ Vc = 3 . 82593,75 = 247781,25 N
Syarat tulangan geser : Ø Vc Vu 3Ø Vc : 82593,75 N 136071,0 N 247781,25 N
Jadi diperlukan tulangan geser Ø Vs
= Vu – Ø Vc = 136071,0 - 82593,75
= 53477,25 N Vs perlu
= 75
, 53477,25
75 ,
= Vs
φ = 71303 N
Av = 2 . ¼
π 10
2
= 2 . ¼ . 3,14 . 100 = 157 mm
2
S = 17
, 349
71303 5
, 440
. 360
. 157
perlu Vs
d .
fy .
Av =
= mm
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500
300
4 D19
Ø10-200 2 D19
500
300
2 D19
Ø10 -200 4 D19
Tul. Tumpuan Tul. Lapangan
S max = d2 = 2
5 ,
440 = 220,25 mm
Jadi dipakai sengkang dengan tulangan Ø 10 – 200 mm
Potongan portal melintang
7.8. Penulangan Kolom
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